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  • 1 answers

Aviral Goyal 6 years ago

Bro tanA should be equal to 3
  • 2 answers

Komal ...? 6 years ago

Bcz of chnged pttrn..and also acc. to basic nd standard..

Sumit Kumar 6 years ago

The pattern has changed
2x-
  • 1 answers

Komal ...? 6 years ago

2x- ..?? Question is not clear..
  • 1 answers

Raunak Chauhan 6 years ago

Munna bbhukha lahi thi kya jo addha se jada andra ho gaya . Take care again during writing . Best of luck for your board examintion
  • 1 answers

Raunak Chauhan 6 years ago

HCF of 210&55 is 5 According to the question 5=210*5+55y 5=1050+55y -1045=55y Y=-1045/55 Y=-209/11
  • 2 answers

Jatin Rajput 6 years ago

Yes

Shivam Tiwari 6 years ago

Theorem 6.1 and 6.2
  • 3 answers

Ram ? 6 years ago

1 by 4

Yogita Ingle 6 years ago

When two different coins are tossed randomly, the sample space is given by

S = {HH, HT, TH, TT}

Therefore, n(S) = 4.

Getting two heads:

Let E1 = event of getting 2 heads. Then,
E1 = {HH} and, therefore, n(E1) = 1.
Therefore, P(getting 2 heads) = P(E1) = n(E1)/n(S) = 1/4.

Prafful Singhal 6 years ago

3/4
  • 1 answers

Vijendra Gupta 6 years ago

2sinAcosA = 2sinA CosA=1 cosA = cos0 A = 0
  • 1 answers

Vijendra Gupta 6 years ago

Let asinA - bcosA =x Square both side both equations and add them. Using identity sin2A + cos2A = 1 We get a2 + b2 = c2 + x2 x = ✓a2 + b2 - c2
  • 2 answers

Abhinav ? 6 years ago

<A><FONT size="3">All the best you too bro..

Aadya Singh ? 6 years ago

<I><font size = "4cm" Color ="blue"> Thanku....?
  • 3 answers

Aadya Singh ? 6 years ago

Maths...
Ok

Aditi Nagesh 6 years ago

I will choose maths with additional biology
  • 5 answers

Deep Singh 6 years ago

p(x) = x^2 -7x- 10

Deep Singh 6 years ago

x^2-(α+β)+αβ = x^2 -7 +10 Jo isse phle answer ki thi wo glt solve kr di thi....yeh solution shi h dekh lo
Because sum of zero is given 7 means (alpha + bita )=7.and (alpha×bita)=10 that is product of zero
X^2 -7x+10

Deep Singh 6 years ago

x^2-(α+β)x+αβ X^2-(7+10)x+(7*10) X^2-17x+70 Therefore,p(x)=x^2-17x+70.
  • 2 answers
X(1×2×8) x in common =16x

Deep Singh 6 years ago

Kay h yeh??
  • 0 answers
  • 2 answers

Deep Singh 6 years ago

1

Yogita Ingle 6 years ago

Numbers we will use

=> 2930 - 7 = 2923

=> 3250 - 11 = 3239

Thus by Prime Factorisation Method

=> 2923 = 37 x 79

=> 3239 = 41 x 79

Thus HCF(2923,3239) = 79

Thus, the greatest number that will divide 2930 and 3250 to leave 7 and 11 as remainder.

  • 1 answers
Wooo what a question

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